原题地址：http://oj.leetcode.com/problems/reorder-list/

题意：

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes’ values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

解题思路：1，先将链表截断为两个相等长度的链表，如果链表长度为奇数，则第一条链表长度多1。如原链表为L={1,2,3,4,5}，那么拆分结果为L1={1,2,3}；L2={4,5}。拆分的技巧还是快慢指针的技巧。

　　　　   2，将第二条链表L2翻转，如将L2={4,5}翻转为L2={5,4}。

              3，按照题意归并链表。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @return nothing
    def reorderList(self, head):
        if head==None or head.next==None or head.next.next==None: return head
        
        # break linked list into two equal length
        slow = fast = head                              #快慢指针技巧
        while fast and fast.next:                       #需要熟练掌握
            slow = slow.next                            #链表操作中常用
            fast = fast.next.next
        head1 = head
        head2 = slow.next
        slow.next = None

        # reverse linked list head2
        dummy=ListNode(0); dummy.next=head2             #翻转前加一个头结点
        p=head2.next; head2.next=None                   #将p指向的节点一个一个插入到dummy后面
        while p:                                        #就完成了链表的翻转
            tmp=p; p=p.next                             #运行时注意去掉中文注释
            tmp.next=dummy.next
            dummy.next=tmp
        head2=dummy.next

        # merge two linked list head1 and head2
        p1 = head1; p2 = head2
        while p2:
            tmp1 = p1.next; tmp2 = p2.next
            p1.next = p2; p2.next = tmp1
            p1 = tmp1; p2 = tmp2